Define a multiplicative function $r(n)$ by $r(p^a)=\frac{\binom{2a}{a}}{4^a}$. Show that $(\sum_{n\leq p^{3/2}}\frac{r(n)}{n})^2<< \log p$

Question

Define a multiplicative function $r(n)$ by $r(p^a)=\frac{\binom{2a}{a}}{4^a}$. Show that $$\left(\sum_{n\leq p^{3/2}}\frac{r(n)}{n}\right)^2<< \log p$$ I am confused about, how $\log(p)$ is coming here. Can we use Euler's summation formula to get the bound?

Answer 1

${-1/2 \choose a+1}= {-1/2 \choose a}\frac{-1/2-a}{a+1}$, $ {2(a+1) \choose (a+1)}={2a \choose a} \frac{(2a+1)(2a+2)}{(a+1)^2}={2a \choose a}\frac{4(a+1/2)}{a+1}= 4^{a+1}(-1)^{a+1}{-1/2 \choose a+1} $

$$1+\sum_{a=1}^\infty r(p^a) x^a = \sum_{a=0}^\infty {2a \choose a} 4^{-a} x^a = \sum_{a=0}^\infty {-1/2 \choose a}(-1)^a x^a = (1-x)^{-1/2}$$

thus $$(\sum_{n \le x} r(n) n^{-1})^2 \le (\prod_{p \le x} (1+\sum_{a=1}^\infty r(p^a) p^{-a}))^2 = (\prod_{p \le x} (1-p^{-1})^{-1/2})^2 = \prod_{p \le x} (1-p^{-1})^{-1} \\ = \exp(-\sum_{p \le x}\log(1-p^{-1})) \le C \exp(\log \log x) \le C \log x$$ Where $-\sum_{p \le x}\log(1-p^{-1}) \sim \log \log x$ is Mertens's theorem.